A person jumps from the roof of a house 4.0- high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.74. 1)If the mass of his torso (excluding legs) is 40 , find his velocity just before his feet strike the ground. 2)If the mass of his torso (excluding legs) is 40 , find the magnitude of the average force exerted on his torso by his legs during deceleration.

mgh=1/2 m v^2

v=sqrt 4*g

2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74

check that.

I don't understand this part....

2) force*distance=1/2 m v^2
force= mass*(4g)/2d= 2*g*mass/.74

To solve this problem, we will use the principles of conservation of energy and Newton's second law.

1) Finding the velocity just before his feet strike the ground:
We can use the principle of conservation of energy to find the initial velocity of the person just before his feet strike the ground.

The potential energy gained from falling from a height of 4.0 m is given by:
PE = m * g * h
where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

So, the potential energy gained is:
PE = 40 * 9.8 * 4.0

This potential energy gets converted to kinetic energy just before the person's feet strike the ground:
KE = (1/2) * m * v^2
where v is the velocity just before his feet strike the ground.

So, we have:
PE = KE
40 * 9.8 * 4.0 = (1/2) * 40 * v^2

Simplifying this equation, we get:
v^2 = 78.4
v ≈ 8.85 m/s

Therefore, the velocity just before his feet strike the ground is approximately 8.85 m/s.

2) Finding the magnitude of the average force exerted on his torso by his legs during deceleration:
We can use Newton's second law to find the average force exerted on his torso during deceleration.

The force can be calculated using the equation:
F = m * a,
where m is the mass of the torso and a is the deceleration.

The deceleration can be calculated using the equation:
a = (vf^2 - vi^2) / (2 * d),
where vf is the final velocity (0 m/s in this case) and d is the distance over which deceleration occurs (0.74 m in this case). vi is the initial velocity, which is the velocity just before the person's feet strike the ground (8.85 m/s in this case).

Substituting the values, we get:
a = (0^2 - 8.85^2) / (2 * 0.74)

Simplifying this equation, we get:
a ≈ -66.91 m/s^2

Finally, substituting the values of mass and deceleration into Newton's second law equation, we get:
F = 40 * (-66.91)
F ≈ -2676.4 N

Therefore, the magnitude of the average force exerted on his torso by his legs during deceleration is approximately 2676.4 N (positive because force is a vector quantity, and we are only considering the magnitude here).

To answer these questions, we can use the principles of conservation of energy and Newton's second law of motion.

1) Let's determine the velocity just before the person's feet strike the ground.

First, we need to calculate the potential energy of the person when they are on the roof. The potential energy is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

PE = 40 kg * 9.8 m/s^2 * 4.0 m
PE = 1568 J

Next, we need to calculate the kinetic energy of the person just before their feet hit the ground. The kinetic energy is given by the formula: KE = 0.5mv^2, where v is the velocity.

KE = 0.5 * 40 kg * v^2

According to the principle of conservation of energy, the potential energy at the roof is equal to the kinetic energy just before striking the ground:

PE = KE
1568 J = 0.5 * 40 kg * v^2

Solving for v:

v^2 = (2 * 1568 J) / (40 kg)
v^2 = 78.4 m^2/s^2
v = √(78.4 m^2/s^2)
v ≈ 8.85 m/s

Therefore, the velocity just before the person's feet strike the ground is approximately 8.85 m/s.

2) Now, let's calculate the magnitude of the average force exerted on the person's torso by their legs during deceleration.

Using Newton's second law, we know that force (F) equals mass (m) multiplied by acceleration (a): F = ma.

To find the acceleration, we can use the formula: v^2 = u^2 + 2as, where u is the initial velocity, a is the acceleration, s is the displacement.

Here, the person's initial velocity (u) is 8.85 m/s, the final velocity (v) is 0 m/s, and the displacement (s) is 0.74 m.

v^2 = u^2 + 2as
(0 m/s)^2 = (8.85 m/s)^2 + 2a * 0.74 m

Simplifying the equation:

0 = 78.3225 m^2/s^2 + 1.48a m

Rearranging the equation to solve for acceleration (a):

a = -78.3225 m^2/s^2 / 1.48 m
a ≈ -52.89 m/s^2

Now that we have the acceleration, we can calculate the magnitude of the force:

F = m * a
F = 40 kg * 52.89 m/s^2
F ≈ 2116 N

Therefore, the magnitude of the average force exerted on the person's torso by their legs during deceleration is approximately 2116 N.